Question

If \( 0 \le A, B \le \frac{\pi}{4} \) and \( \cot A + \cot B + \tan A + \tan B = \cot A \cot B - \tan A \tan B \) then \( \sin(A+B) = \)

• A. 0
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{\sqrt{2}} \)
• D. \( \frac{\sqrt{3}}{2} \)

Hint:

When an equation involves a mix of all four trigonometric functions (sin, cos, tan, cot), converting everything to sine and cosine is a reliable strategy. Look for opportunities to apply sum and difference formulas like \( \sin(A\pm B) \) and \( \cos(A\pm B) \).

Answer 1

The Correct Option is C

Start by rewriting the given equation entirely in terms of sine and cosine.
\( \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\sin A}{\cos A} + \frac{\sin B}{\cos B} = \frac{\cos A \cos B}{\sin A \sin B} - \frac{\sin A \sin B}{\cos A \cos B} \).
Combine the first two terms and the next two terms on the LHS.
\( \frac{\cos A \sin B + \sin A \cos B}{\sin A \sin B} + \frac{\sin A \cos B + \cos A \sin B}{\cos A \cos B} = \frac{\cos^2 A \cos^2 B - \sin^2 A \sin^2 B}{\sin A \sin B \cos A \cos B} \).
The numerator in both terms on the LHS is \( \sin(A+B) \).
\( \frac{\sin(A+B)}{\sin A \sin B} + \frac{\sin(A+B)}{\cos A \cos B} = \frac{(\cos A \cos B - \sin A \sin B)(\cos A \cos B + \sin A \sin B)}{\sin A \sin B \cos A \cos B} \).
Factor out \( \sin(A+B) \) on the LHS. The RHS numerator is \( \cos(A+B)\cos(A-B) \).
\( \sin(A+B) \left( \frac{1}{\sin A \sin B} + \frac{1}{\cos A \cos B} \right) = \frac{\cos(A+B)\cos(A-B)}{\sin A \sin B \cos A \cos B} \).
Combine the terms in the parenthesis on the LHS.
\( \sin(A+B) \left( \frac{\cos A \cos B + \sin A \sin B}{\sin A \sin B \cos A \cos B} \right) = \frac{\cos(A+B)\cos(A-B)}{\sin A \sin B \cos A \cos B} \).
The numerator in the parenthesis is \( \cos(A-B) \).
\( \sin(A+B) \frac{\cos(A-B)}{\sin A \sin B \cos A \cos B} = \frac{\cos(A+B)\cos(A-B)}{\sin A \sin B \cos A \cos B} \).
Since \( 0 \le A, B \le \pi/4 \), \( A-B \) is in \( [-\pi/4, \pi/4] \), so \( \cos(A-B) \neq 0 \). We can cancel it.
This leaves us with \( \sin(A+B) = \cos(A+B) \).
This implies \( \tan(A+B) = 1 \).
Given the range \( 0 \le A, B \le \pi/4 \), we have \( 0 \le A+B \le \pi/2 \).
In this interval, \( \tan(A+B)=1 \) means \( A+B = \frac{\pi}{4} \).
We are asked to find \( \sin(A+B) \), which is \( \sin(\frac{\pi}{4}) \).
\( \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \).